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How to show $z=\pm\pi$ is a removable singularity for $\frac1{\sin z}+\frac{2z}{z^2-\pi^2}$? I tried to compute the Laurent series, specifically the coefficients for $1/z,1/z^2,...$ since if we can show those coefficients are all zeroes, we are done. Is there a nice way to compute those coefficients? I tried the contour integral formula but couldn't get anywhere with it. Thanks. By the way the latter term can be written as $\frac1{z+\pi}+\frac1{z-\pi}$.

Another related question: how to show $z=0$ is a removable singularity for $\frac1{\sin z}-1/z$?. I tried to write it as $z\left(\frac{z/3!-z^3/5!+z^5/7!-z^7/9!+...}{z-z^3/3!+z^5/5!-...}\right)$ using the Maclaurin series for $\sin z$. But how to show rigorously that the expression can be written as $z\sum_{k=0}^{\infty}a_kz^k$?

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Related problems: (I), (II). Just take the limit as $z\to \pi$ and $z\to -\pi$ and the two limits should be finite. For instance

$$ \lim_{z\to \pi} \left(\frac1{\sin z}+\frac{2z}{z^2-\pi^2}\right)=\lim_{z\to \pi} \frac{(z^2-\pi^2)+2z\sin(z)}{(z^2-\pi^2)\sin(z)} = \frac{1}{2\pi}. $$

It is just like the function $\frac{\sin(z)}{z}$ which has $z=0$ as a removable singularity, since

$$ \lim_{z\to 0}\frac{\sin(z)}{z}=1. $$

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  • $\begingroup$ How did you get $\frac1{2\pi}$? $\endgroup$ – user70520 Apr 3 '13 at 4:42
  • $\begingroup$ @user70520: Simplify the expression and then use L'hobital rule (since you will have the form $\frac{0}{0}$). I have to leave you now. It is getting late. $\endgroup$ – Mhenni Benghorbal Apr 3 '13 at 4:46
  • $\begingroup$ @user70520: Have you benefited of this answer? Note that, it is a technique that does not depend on deriving Laurent series. It is easier and a short cut. $\endgroup$ – Mhenni Benghorbal Apr 3 '13 at 12:05
  • $\begingroup$ Yeah I forgot I can apply L'Hopital's rule in the complex plane. $\endgroup$ – user70520 Apr 4 '13 at 1:39
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Taking $z = \pi + w$ we get (as $z \to \pi$) $$\eqalign{\frac{1}{\sin(z)} &= \frac{1}{-\sin(w)} = \frac{1}{-w + O(1)} = \frac{-1}{w} + O(1)\cr \frac{2z}{z^2 - \pi^2} &= \frac{2(\pi + w)}{w^2 + 2 \pi w} = \frac{1+O(w)}{w+O(w^2)} = \frac{1}{w} + O(1)\cr}$$ Thus $$\frac{1}{\sin(z)} + \frac{2z}{z^2 - \pi^2} = O(1)$$

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